Capacity of MIMO

8/2/2019 Capacity of MIMO

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Capacity of MIMO channels

Nghi H. Tran

Supervisor: Prof. Ha H. Nguyen

Department of Electrical Engineering

University of Saskatchewan

Nghi H. Tran, UoS Capacity of MIMO channels 1


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Outline

Channel Model

Mathematical Preniminaries

Channel with fixed transfer function

Channel with Rayleigh fading

Example with Orthogonal Design

Non-ergodic channel



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System Model

We deal with a linear model with the received vector y Cr andthe transmitted vector x Ct:

y = Hx + n (1)

with H is a r t complex matrix and n is complex Gaussian

noise with independent components, E[nn] = Ir.

The power constraint: E[xx] P, or equivalently

tr(E[xx]) P. So does for x E(x) interested inzero-mean x.

Three cases ofH:

Deterministic

H is random, chosen according to a pdf, and each use of the

channel corresponds to an independent realization.H is random, but is fixed once it is chosen.



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Mathematical Preliminaries

A complex vector x Cn is Gaussian ifx =

Re(x)

Im(x)

is Gaussian.

To specify the distribution ofx, it is necessary to specify E[x] andE

(x E[x])(x E[x])

A complex Gaussian x is circularly symmetric if:

E

(x E[x])(x E[x])

=1

2

Re(Q) Im(Q)Im(Q) Re(Q)

(2)for some Hermitain non-negative definite Q.

For circularly symmetric x, the first and the second statistics can be specified by

E[x] and E

(x E[x])(x E[x])

In Tse book: Circular symmetric: x and exp(j)x has the same distribution. Itleads to the mean = 0



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The density function of a circularly symmetric complex Gaussian

x with mean and covariance Q is given as:

,Q(x) = det(Q)1 exp (x )Q1(x ) (3)

The differential entropy ofx is then computed as:

H(Q) = EQ [ log Q(x)]

= log det(Q) + (log(e))E[xQ1x]

= log det(Q) + (log(e))tr(E[xx]Q1

)

= log det(eQ) (4)



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The importance of the circularly symmetric complex Gaussian is

due to the fact that, given Q, x maximize the entropy.

How to prove: Using following inequality:

Cnp(x)logp(x)dx Cnp(x)logpN(x)dx (5)where p(x) and pN(x) are arbitrary pdfs. After that, apply it by

using pN(x) = Q(x).

Ifx Cn is a circularly symmetric complex Gaussian then so is

y = Ax for any A Cmn

Ifx and y are independent circularly symmetric complexGaussian, then so is z = x + y.



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Channel With Fixed H

Using singular value decomposition theorem:

H = UDV (6)

where U Crr and V Ctt are unitary, and D Rrt is non-negative and

diagonal.

In fact, the diagonal entries ofD are the singular values ofH, the columns ofU

and U are the eigenvectors ofHH and HH, respectively.

Then we have:

y = UDVx+ n (7)

The fact thatUn and n has the same distribution, we have an equivalent channel:

y = Dx+ n (8)

where n = Un, y = Uy and x = Vx



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Channel With Fixed H

Since rank(H) is at most min{r, t}, at most min{r, t} of its singular values are

non-zero, denoted by 1/2i

Therefore, we have parallel Gaussian channels:

yi = 1/2i xi + ni, 1 i min{r, t} (9)

Clearly, yi, i > min{r, t} is just noise component and xi does not play any role.

The well-known result: {xi i min{r, t}} to be independent Gaussians.

The variances ofxi are chosen via water filling" as:

E[Re(xi)2

] = E[Im(xi)2

] =

1

2 ( 1

i )+

(10)

where is chosen to meet the power constraint P() =

i( 1i )

+

The maximal mutual information:

C() =i

(ln(i))+

(11)



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Alternative Derivation

The mutual information I(x,y) is given:

I(x,y) = H(y) H(n) (12)

The covariance ofy: E[yy] = HQH + Ir

To maximize H(y) with given Q, x should be circularly

symmetric Gaussian.Hence, we have:

I(x,y) = log det(Ir + HQH) = log det(It + QH

H) (13)

(Using determinant identity det(Ia + AB) = det(Ib + BA))

Now we need to choose Q subject to tr(Q) P and Q is

non-negative definite.



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Alternative Derivation

Diagonalize HH as follows (it is Hermitian):

HH = UU (14)

with unitary U and = diag(1, . . . , t)

It then follows:det(It +QH

H) = det(It +QUU) = det(It +UQU

)

= det(It + 1/2UQU1/2) (15)

Observe that the properties ofQ = UQU is the same withQ Can maximizeover Q.

Furthermore, for non-negative definite matrixA, detA i Aii. Then we have:det(It + 1/2Q1/2) i

(1 + Qiii) (16)

The equality holds when Q is diagonal. Thus we see that the maximizing Q is

diagonal and the optimal diagonal one can be obtained using water filling.

Results: Qii = ( 1i ) where

i Qii = P.


The Gaussian Channel with Rayleigh


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The Gaussian Channel with Rayleigh

Fading

Consider H is random, independent realization for each block, each entry is

zero-mean, complex Gaussian, variance 1/2, independent component.

Lemma: For any unitary U Crr

and V Ctt

,UHV

,UH,HV and Hhas the same distribution.

The mutual information is given as:

I(x,y) = EH[I(x,y)|H] (17)

For a given covariance Q ofx, I(x,y)|H is maximized when x is circularly

symmetric Gaussian.

Therefore, now, we need to maximize

(Q) = E

log det(Ir +HQH

(18)

over the choice of non-negative definiteQ subject tr(Q) P



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Capacity

Since Q is non-negative definite,Q = UDU with unitaryU and diagonal

non-negative D

Then we have:

(Q) = E

log det(Ir + (HU)D(HU)

(19)

Since H and HU have the same distribution, (Q) = (D) restrict attention

to non-negative diagonalQ

Considering t! permutation matrix corresponding to Q = Q. It is easy to

see (Q) = (Q)

Observe: log det is concave on the set of positive definite matrices. It then

follows that is concave.

(aQ1 + (1 a)Q2) a(Q1) + (1 a)(Q2) (20)

where 0 a 1.Nghi H. Tran, UoS Capacity of MIMO channels 12


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Evaluation of Capacity

The capacity of channel with power constraint P is:

0

log(1+P/t)m1

k=0

k!

(k + n m)! Lnmk ()2

nm exp()d

(23)where m = min{r, t}, n = max{r, t}, Lij: associated Laguerre

polynomials.

At high SNR, we have:

C(SN R) = m log(SN R) + O(1) (24)

where SN R is average signal to noise ratio at each receiveantenna. In this model, SN R = P

Clearly, C increases with m log(SN R), in contrast to log(SN R)

for single antenna. MIMO can be viewed as m parallel spatialchannels.



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Evaluation of Capacity: Low P

Rewrite C based on singular values ofH 1/2i :

C(P) = Em

i=1 log1 +P

ti (25)

At low P, using log2(1 + x) x log2 e:

C

mi=1

P

t E[i]log2 e =

P

t E[tr(HH

)]log2 e

=P

tE

i,j

|hi,j |2 = P.r. log2 e (26)

where using the fact: sum of eigenvalues equal to trace.

Clearly, with low P, multiple transmitter antennas are not veryuseful.



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Example with Alamouti Scheme

For the convenience, to get SNR at each receive antenna is :

y =

tHx+ n (27)

The capacity:

C(,t,r) = E[log det(Ir +

tHH) = E[log det(It +

tHH) (28)

With r = 1, Alamouti Scheme is equivalent to :

y1

y2

=

2

h1 h2

h2 h1

x1

x1

+

n1

n2

(29)

The capacity of this equivalent channel is:

Corth() =1

2

log detI2 +

2

(|h1|2 + |h2|

2)I2 = C(, 2, 1) (30)Clearly, there is no loss in terms of capacity with r = 1.



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Example with Alamouti Scheme

For r = 2, an equivalent channel:

y11

y21

y12

y22

=

2

h11 h21

h21 h11

h12 h22

h22 h12

H

x1x2

+

v11

v21

v12

v22

(31)

With HH =

i,j |hi,j |2I2, we have:

Corth() =1

2E

log det

I2 +

2(|h11|

2 + |h12|2 + |h21|

2 + |h22|2)I2

= Elog1 + 2

4(|h11|2 + |h12|2 + |h21|2 + |h22|2)

= C(2, t = 4, r = 1) < C(, t = 2, n = 2) (32)

It says that Alamouti scheme achieves the capacity oft = 4, r = 1 and at twiceSNR. General result: C(r, 2r, 1) < C(, 2, r).


2 2


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Capacity 2 2 Channel

0 5 10 15 20 25 300

2

4

6

8

10

12

14

16

18

SNR(dB)

C(bits

)

r=2, t=2

Actual 2x2 channelAlamouti scheme


N di Ch l


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Non-ergodic Channels

H is chosen randomly at the beginning and is held fixed for all the

uses of the channel.

In this case, given data rate R, when log det(Ir +HQH) < R, noreliable communication System is in outage.

Outage probability given R and P:

Pout(R, P) = P(log det(Ir + HQH) < R) (33)

Given R, if system is not in outage, there exists a universal codingstrategy that achieves reliable communication

A transmit strategy can be chosen (parameterized by the

covariance) to minimize the probability of outage event.


Capacity of MIMO

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